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ICE Princess25 [194]
3 years ago
7

1. How can you find the decimal forms of 3/12 and 2/9?

Mathematics
1 answer:
Helga [31]3 years ago
6 0

Answer:

1. 0.472

2. 0.25

Step-by-step explanation:

1. reduce the fraction with 3 = 1\4 + 2\9

write all numerators above the Least common denominator 36 =9 + 8\36

Add numbers = 17\36 and the alternate form is 0.472

2. Divide the numerator and denominator by three = 3 ÷ 3\12 ÷ 3 = 0.25

I can't do the others now I have an exam(sorry)

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John has a storage bin in the shape of a rectangular prism the storage bed is still in the and 1/2 ft long 2 ft wide and 2 ft ta
Phoenix [80]

We know, volume of rectangular prism is given by :

V = lbh\\\\V=\dfrac{1}{2}\times 2\times 2\ ft^2\\\\V=2\ ft^2

Volume of cubic box :

v=a^3\\\\v=(\dfrac{1}{2})^3\ ft^3\\\\v=\dfrac{1}{8}\ ft^3

Number of cubes can be filled :

N=\dfrac{V}{v}\\\\N=\dfrac{2}{\dfrac{1}{8}}\\\\N=16

Therefore, 16 cubes can be put in the bin.

Hence, this is the required solution.

8 0
3 years ago
BRAINLIEST ANSWER<br> please help me with this n I'll mark u the brainliest
Artyom0805 [142]
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7 0
3 years ago
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For what values of θ is sinθ &lt; cosθ when 0 ≤ θ &lt; π ?
zysi [14]
<h3>Answer:  0 \le \theta < \frac{\pi}{4}</h3>

=========================================================

How to get this answer:

Use the unit circle to note that \sin\theta = \cos\theta = \frac{\sqrt{2}}{2} when \theta = \frac{\pi}{4} (aka 45 degrees)

Beyond this point, cosine is smaller than sine. This means that anything from 0 to pi/4 will have sine be smaller than cosine. It might help to graph y = sin(x) and y = cos(x) on the interval from x = 0 to x = pi.

The two curves y = sin(x) and y = cos(x) intersect at the point \left(\frac{\pi}{4}, \frac{\sqrt{2}}{2}\right)

-------------------------------

Here's a more detailed picture of whats going on.

\sin \theta < \cos \theta\\\\\sin \theta < \sqrt{1-\sin^2\theta}\\\\\sin^2 \theta < 1-\sin^2\theta \\\\2\sin^2 \theta < 1\\\\\sin^2 \theta < \frac{1}{2}\\\\\sin \theta < \sqrt{\frac{1}{2}}\\\\\sin \theta < \frac{1}{\sqrt{2}}\\\\\sin \theta < \frac{\sqrt{2}}{2}\\\\\theta < \arcsin\left(\frac{\sqrt{2}}{2}\right)\\\\\theta < \frac{\pi}{4}\\\\

Intersect the intervals 0 \le \theta < \pi and \theta < \frac{\pi}{4} and you'll end up with the final answer 0 \le \theta < \frac{\pi}{4}

4 0
3 years ago
Find <br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bdy%7D%7Bdx%7D%20" id="TexFormula1" title=" \frac{dy}{dx} " alt=" \frac{d
nataly862011 [7]

Answer:

\displaystyle y' = 2x + 3\sqrt{x} + 1

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Algebra I</u>

  • Terms/Coefficients
  • Anything to the 0th power is 1
  • Exponential Rule [Rewrite]:                                                                              \displaystyle b^{-m} = \frac{1}{b^m}
  • Exponential Rule [Root Rewrite]:                                                                     \displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}<u> </u>

<u>Calculus</u>

Derivatives

Derivative Notation

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:                                                                                    \displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

<em />\displaystyle y = (x + \sqrt{x})^2<em />

<em />

<u>Step 2: Differentiate</u>

  1. Chain Rule:                                                                                                        \displaystyle y' = 2(x + \sqrt{x})^{2 - 1} \cdot \frac{d}{dx}[x + \sqrt{x}]
  2. Rewrite [Exponential Rule - Root Rewrite]:                                                     \displaystyle y' = 2(x + x^{\frac{1}{2}})^{2 - 1} \cdot \frac{d}{dx}[x + x^{\frac{1}{2}}]
  3. Simplify:                                                                                                             \displaystyle y' = 2(x + x^{\frac{1}{2}}) \cdot \frac{d}{dx}[x + x^{\frac{1}{2}}]
  4. Basic Power Rule:                                                                                             \displaystyle y' = 2(x + x^{\frac{1}{2}}) \cdot (1 \cdot x^{1 - 1} + \frac{1}{2}x^{\frac{1}{2} - 1})
  5. Simplify:                                                                                                             \displaystyle y' = 2(x + x^{\frac{1}{2}}) \cdot (1 + \frac{1}{2}x^{-\frac{1}{2}})
  6. Rewrite [Exponential Rule - Rewrite]:                                                              \displaystyle y' = 2(x + x^{\frac{1}{2}}) \cdot (1 + \frac{1}{2x^{\frac{1}{2}}})
  7. Multiply:                                                                                                             \displaystyle y' = 2[(x + x^{\frac{1}{2}}) + \frac{x + x^{\frac{1}{2}}}{2x^{\frac{1}{2}}}]
  8. [Brackets] Add:                                                                                                 \displaystyle y' = 2(\frac{2x + 3x^{\frac{1}{2}} + 1}{2})
  9. Multiply:                                                                                                             \displaystyle y' = 2x + 3x^{\frac{1}{2}} + 1
  10. Rewrite [Exponential Rule - Root Rewrite]:                                                     \displaystyle y' = 2x + 3\sqrt{x} + 1

Topic: AP Calculus AB/BC (Calculus I/II)

Unit: Derivatives

Book: College Calculus 10e

4 0
3 years ago
Josh ran 84.6 miles in June. In July he ran 0.55 times this distance due to an injury. How many more miles did he run in June th
Irina-Kira [14]

I did 0.80-0.25=0.55

so I figured that is what she had left so I know took the total 0.80-0.55 and got0.25 which is B.

My friend said the correct answer was A.

Don't worry I just did this problem!

5 0
2 years ago
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